Proof of Coordinate Functions, the $C^k$ Hierarchy, and Partial Derivatives of a Smooth Map

Throughout, "clause 1", "clause 2" and "clause 3" refer to the corresponding clauses of C^k Maps on a Euclidean Open Set. We use repeatedly that, by clause 3, a function $g:U\to\mathbb{R}$ is of class $C^{k}$ on $U$ exactly when the map into $\mathbb{R}^{1}$ with single coordinate function $g$ is, so that clauses 1 and 2 read for $g$ with the index $j$ suppressed. We also use that every natural number is either $1$ or of the form $k+1$ for a natural number $k$: the set of natural numbers of one of these two forms contains $1$ and contains $k+1$ whenever it contains $k$, hence is all of the natural numbers by induction.

Claim 1. First let $k=1$. By clause 1, $F$ is of class $C^{1}$ on $U$ if and only if for every $j$ with $1\le j\le m$ the coordinate function $F_j$ is continuous at every point of $U$ and, for every $i$ with $1\le i\le n$, the partial derivative of $F_j$ with respect to the $i$th variable exists at every point of $U$ and $\partial_i F_j$ is continuous at every point of $U$. For a fixed $j$, clause 1 applied to $F_j$ in the sense of clause 3 says that $F_j$ is of class $C^{1}$ on $U$ if and only if those same conditions hold for that $j$. Hence $F$ is of class $C^{1}$ on $U$ if and only if every $F_j$ is.

Now let $k$ be a natural number and consider class $C^{k+1}$. By clause 2, $F$ is of class $C^{k+1}$ on $U$ if and only if $F$ is of class $C^{1}$ on $U$ and $\partial_i F_j$ is of class $C^{k}$ on $U$ for all $i$ and $j$ with $1\le i\le n$ and $1\le j\le m$; and, for fixed $j$, $F_j$ is of class $C^{k+1}$ on $U$ if and only if $F_j$ is of class $C^{1}$ on $U$ and $\partial_i F_j$ is of class $C^{k}$ on $U$ for every such $i$. Combining these with the case $k=1$ already proved, $F$ is of class $C^{k+1}$ on $U$ if and only if, for every $j$, $F_j$ is of class $C^{1}$ on $U$ and $\partial_i F_j$ is of class $C^{k}$ on $U$ for every $i$; that is, if and only if every $F_j$ is of class $C^{k+1}$ on $U$. This proves the first assertion for every natural number $k$.

For the second assertion, $F$ is smooth on $U$ if and only if $F$ is of class $C^{k}$ on $U$ for every natural number $k$, and the same criterion applies to each $F_j$, whose smoothness is defined through the scalar convention of clause 3. By the first assertion, $F$ is of class $C^{k}$ on $U$ for every $k$ if and only if every $F_j$ is of class $C^{k}$ on $U$ for every $k$; that is, if and only if every $F_j$ is smooth on $U$.

Claim 2. By claim 1 it suffices to prove the assertion for functions $g:U\to\mathbb{R}$: if it holds for those, then $F$ of class $C^{k+1}$ on $U$ makes every $F_j$ of class $C^{k+1}$, hence of class $C^{k}$, hence makes $F$ of class $C^{k}$. So we prove by induction on the natural number $k$ the statement: every $g:U\to\mathbb{R}$ of class $C^{k+1}$ on $U$ is of class $C^{k}$ on $U$.

For $k=1$: if $g$ is of class $C^{2}$ on $U$, then $g$ is of class $C^{1}$ on $U$ by clause 2 applied with $k=1$. Assume the statement for $k$, and let $g$ be of class $C^{k+2}$ on $U$. By clause 2, applied with $k+1$ in place of $k$, $g$ is of class $C^{1}$ on $U$ and $\partial_i g$ is of class $C^{k+1}$ on $U$ for every $i$ with $1\le i\le n$. By the inductive hypothesis applied to each $\partial_i g$, each $\partial_i g$ is of class $C^{k}$ on $U$. By clause 2 again, $g$ is of class $C^{k+1}$ on $U$. This completes the induction.

Claim 3. Let $F$ be smooth on $U$ and fix $i$ and $j$ with $1\le i\le n$ and $1\le j\le m$. By claim 1, $F_j$ is smooth on $U$; in particular $F_j$ is of class $C^{1}$ on $U$, so by clause 1 the partial derivative of $F_j$ with respect to the $i$th variable exists at every point of $U$ and $\partial_i F_j$ is a function from $U$ to $\mathbb{R}$. Let $k$ be a natural number. Since $F_j$ is smooth on $U$, it is of class $C^{k+1}$ on $U$, so clause 2 gives that $\partial_i F_j$ is of class $C^{k}$ on $U$. As $k$ was an arbitrary natural number, $\partial_i F_j$ is smooth on $U$.